Video Transcript
In this video, we’ll learn how to
find missing lengths in a triangle containing two or three parallel lines using
proportionality. We’ll be able to recognize when
parallel lines inside or outside triangles create similar triangles, sometimes
called the side splitter theorem, use proportionality to find unknown lengths in
triangles with parallel lines, and we’ll also look at the inverse of the side
splitter theorem.
Let’s start by recalling some
properties of parallel lines. We know, for example, that when two
parallel lines are cut by a transversal, the resulting corresponding angles are
equal. We know also that by adding a
second transversal, we can form two triangles. Now, giving each vertex a label, we
can define the larger triangle 𝐴𝐷𝐸 and the smaller triangle 𝐴𝐵𝐶.
Now, since the two pairs of
corresponding angles are equal, we can say that triangle 𝐴𝐷𝐸 is similar to
triangle 𝐴𝐵𝐶. And since these triangles are
similar, the ratios of their corresponding side lengths must be equal. This means 𝐴𝐵 over 𝐴𝐷, that’s
the ratio of 𝐴𝐵 to 𝐴𝐷, is equal to 𝐴𝐶 over 𝐴𝐸, which in turn is equal to
𝐵𝐶 over 𝐷𝐸. So 𝐴𝐵 is to 𝐴𝐷 as 𝐴𝐶 is to
𝐴𝐸, and 𝐴𝐶 is to 𝐴𝐸 as 𝐵𝐶 is to 𝐷𝐸.
We’re going to use these properties
of similar triangles in our first example to identify which pair of side lengths
have equal proportions when a triangle is cut by a line parallel to one of its
sides.
Using the diagram, which of the
following is equal to 𝐴𝐵 over 𝐴𝐷. Is it (A) 𝐴𝐶 over 𝐸𝐶 or (B)
𝐴𝐵 over 𝐷𝐵 or (C) 𝐴𝐷 over 𝐷𝐵, answer (D), 𝐴𝐶 over 𝐴𝐸, or (E) 𝐴𝐸
over 𝐸𝐶.
We see from the diagram that
the base of triangle 𝐴𝐸𝐷, that′s side 𝐸𝐷, is parallel to the base of
triangle 𝐴𝐵𝐶, that′s side 𝐶𝐵. Since corresponding angles must
be equal if the two lines are parallel, angle 𝐷𝐸𝐴 is equal to angle 𝐵𝐶𝐴,
and angle 𝐸𝐷𝐴 is equal to angle 𝐶𝐵𝐴. Then side 𝐸𝐷 creates triangle
𝐴𝐷𝐸, which is similar to the larger triangle 𝐴𝐵𝐶. Since these triangles are
similar, the ratios of their corresponding side lengths must be equal. In particular, 𝐴𝐸 over 𝐴𝐶
is equal to 𝐴𝐷 over 𝐴𝐵.
Now we want to find which of
the given fractions is equal to 𝐴𝐵 over 𝐴𝐷. And we can do this by finding
the reciprocal of both sides of our equation, which gives us 𝐴𝐶 over 𝐴𝐸 is
equal to 𝐴𝐵 over 𝐴𝐷. Hence, 𝐴𝐵 over 𝐴𝐷 is equal
to 𝐴𝐶 over 𝐴𝐸, which corresponds to the given option (D).
In our next example, we see how to
find an unknown length in a triangle using proportions.
Find the value of 𝑥.
We see from the figure that the
lines 𝐴𝐶 and 𝐴𝐵 are transversals that intersect parallel lines 𝐷𝐸 and
𝐵𝐶. And we know that the two pairs
of corresponding angles created by this intersection are equal. That’s angles 𝐷𝐸𝐴 and 𝐵𝐶𝐴
and angles 𝐸𝐷𝐴 and 𝐶𝐵𝐴. This being the case, we can say
the triangles 𝐴𝐵𝐶 and 𝐴𝐷𝐸 are similar triangles since they each have the
common angle 𝐵𝐴𝐶 and their other two angles are also equal.
Now recall that when two
triangles are similar, the ratios of the length of their corresponding sides are
equal. In particular, 𝐴𝐷 is to 𝐴𝐵
as 𝐷𝐸 is to 𝐵𝐶. In other words, 𝐴𝐷 over 𝐴𝐵
is equal to 𝐷𝐸 over 𝐵𝐶. Now, we know that 𝐴𝐷 is equal
to 10 units. 𝐴𝐵 is equal to 10 plus 11
units, that’s 𝐴𝐷 plus 𝐷𝐵. 𝐷𝐸 is equal to 10 units, and
𝐵𝐶 is 𝑥. And so we have 10 over 10 plus
11 is equal to 10 over 𝑥. That is, 10 over 21 is equal to
10 over 𝑥.
Now, solving for 𝑥, we
multiply through by 21𝑥 and divide both sides by 10, and we have 𝑥 equal to
21. Hence, using the given diagram,
we find that 𝑥 is equal to 21 units.
In the two previous examples, we
saw that if a line intersecting two sides of a triangle is parallel to the third
side, then the smaller triangle created by this line is similar to the original
triangle. And since triangles 𝐴𝐵𝐶 and
𝐴𝐷𝐸 are similar, we have that the proportions 𝐴𝐵 over 𝐴𝐷 and 𝐴𝐶 over 𝐴𝐸
are equal. From the diagram, we can see also
that the line segment 𝐴𝐷 can be split into 𝐴𝐵 plus 𝐵𝐷 and the line segment
𝐴𝐸 can be split into 𝐴𝐶 plus 𝐶𝐸.
Now substituting these expressions
into our equation, we have 𝐴𝐵 over 𝐴𝐵 plus 𝐵𝐷 is equal to 𝐴𝐶 over 𝐴𝐶 plus
𝐶𝐸. And we can rearrange this to give
𝐴𝐵 times 𝐴𝐶 plus 𝐶𝐸 is equal to 𝐴𝐶 times 𝐴𝐵 plus 𝐵𝐷. If we next distribute our
parentheses and subtract 𝐴𝐵 times 𝐴𝐶 from both sides, we have 𝐴𝐵 times 𝐶𝐸 is
equal to 𝐴𝐶 times 𝐵𝐷. And we can rearrange this to get
the equal proportions 𝐴𝐵 over 𝐵𝐷 is equal to 𝐴𝐶 over 𝐶𝐸.
This leads us to the side splitter
theorem linking the line segments created when a parallel side is added to a
triangle. The side splitter theorem says that
if a line parallel to one side of a triangle intersects the other two sides of the
triangle, then the line divides those sides proportionally.
In our diagram, the side 𝐵𝐶 is
parallel to the side 𝐷𝐸. So we see that in our case the side
splitter theorem tells us that 𝐴𝐵 is to 𝐵𝐷 as 𝐴𝐶 is to 𝐶𝐸. Note that this can be extended to
include parallel lines outside the triangle. We can form a similar triangle
outside the first triangle with a parallel line as shown in the diagram. And we can deduce an analog of the
side splitter theorem directly from these similar triangles.
In our next example, we’ll see how
to use the side splitter theorem to identify proportional segments of the sides of
triangles so we can calculate a missing length.
In the figure, the sides 𝑋𝑌
and 𝐵𝐶 are parallel. If 𝐴𝑋 equals 18, 𝑋𝐵 equals
24, and 𝐴𝑌 equals 27, what is the length of 𝑌𝐶?
We’re given the side lengths
𝐴𝑋, 𝑋𝐵, and 𝐴𝑌. And we want to find the length
of 𝑌𝐶. We’re also told that the sides
𝑋𝑌 and 𝐵𝐶 are parallel. Now, the side splitter theorem
tells us that if a line parallel to one side of a triangle intersects the other
two sides, then that line divides those two sides proportionally. In particular, in our case this
means that 𝐴𝑌 is to 𝑌𝐶 as 𝐴𝑋 is to 𝑋𝐵. Now, if we substitute our known
lengths into this equation, 𝐴𝑌 is 27, 𝐴𝑋 is 18, and 𝑋𝐵 is 24, we have 27
over 𝑌𝐶 is equal to 18 over 24. We can rearrange this to get
𝑌𝐶 equal to 24 over 18 times 27, which evaluates to 36.
And so using the side splitter
theorem, we find that the length of 𝑌𝐶 is 36 units.
In our next example, we’ll use the
side splitter theorem to help us solve a multistep problem involving triangles and
parallel lines.
The given figure shows a
triangle 𝐴𝐵𝐶. (1) work out the value of
𝑥. And (2) work out the value of
𝑦.
We’re given a triangle with a
line parallel to one side inscribed within it and various lengths of segments of
the sides of the triangle. And we’re asked to find the
value of 𝑥 and the value of 𝑦.
Let’s begin with part (1),
which is finding 𝑥, where we see the two of our side segments involve 𝑥. We note first that a line of
length two units inside the triangle is parallel to the side 𝐵𝐶. Now, the side splitter theorem
tells us that this line divides the two sides 𝐴𝐶 and 𝐴𝐵 proportionally.
Remember, according to the side
splitter theorem, if a line parallel to one side of a triangle intersects the
other two sides of the triangle, then the line divides those sides
proportionally. If we label this line segment
𝐷𝐸 in our diagram, we could say that 𝐴𝐷 over 𝐷𝐵 is equal to 𝐴𝐸 over
𝐸𝐶. Substituting in the given
lengths, we can form an equation which we can solve for 𝑥. That’s three over two 𝑥 plus
three is equal to two over 𝑥 plus five. Now, multiplying both sides by
𝑥 plus five and two 𝑥 plus three and distributing the parentheses, we have
three 𝑥 plus 15 equals four 𝑥 plus six. Subtracting three 𝑥 and six
from both sides and swapping sides, we then have 𝑥 is equal to nine.
So now, making a note of this
and making some space, we can use this value of 𝑥 in part (2) of the question
to find the value of 𝑦. Now, since the two triangles
created by the intersection of side 𝐷𝐸 share a common angle 𝐴 and the pairs
of corresponding angles also created by this line are equal, we can say the
triangles 𝐴𝐷𝐸 and 𝐴𝐵𝐶 are similar triangles. In particular, this means that
the proportions 𝐴𝐷 over 𝐴𝐵 and 𝐷𝐸 over 𝐵𝐶 are equal. Now, we know that 𝐴𝐵 is equal
to the sum 𝐴𝐷 plus 𝐷𝐵. And that’s three plus two 𝑥
plus three. And we have 𝑥 equal to nine
from part (1). And so this evaluates to
24.
So now, substituting our values
𝐴𝐷 equals three, 𝐴𝐵 equals 24, 𝐷𝐸 equals two, and 𝐵𝐶 equals 𝑦 into our
equation, we have three over 24 is equal to two over 𝑦. Now, multiplying both sides by
𝑦 and by 24 over three, we have 𝑦 equal to 16. Hence, from the given figure,
we find that 𝑥 is equal to nine and 𝑦 is equal to 16.
So far, we’ve used the side
splitter theorem to solve problems of proportionality and triangles. We’ve also learned that this can be
extended to parallel lines outside a given triangle. In fact, it turns out that the
converse of this result is also true. And this can be very useful when
solving problems of this type. The converse of the side splitter
theorem tells us that if a line intersecting two sides of a triangle splits these
sides into equal proportions, then that line must be parallel to the third side of
the triangle.
In all three of the diagrams shown,
𝐴𝐵𝐶 is a triangle and the line 𝐷𝐸 intersects 𝐴𝐵 at 𝐷 and 𝐴𝐶 at 𝐸. In other words, if proportions 𝐴𝐷
over 𝐷𝐵 and 𝐴𝐸 over 𝐸𝐶 are equal, then the lines 𝐷𝐸 and 𝐵𝐶 must be
parallel. So let’s now use the converse of
the side splitter theorem to find unknown lengths in a given triangle.
Given that 𝐴𝐵𝐶𝐷 is a
parallelogram, find the length of 𝑌𝑍.
To find the length of the side
𝑌𝑍 of the smaller triangle 𝑋𝑌𝑍 inside the parallelogram, let’s begin by
identifying some relevant information about the two triangles 𝑋𝑌𝑍 and
𝑋𝐷𝐶. We’re given that 𝑋𝑌 is equal
to 𝑌𝐷 and 𝑋𝑍 is equal to 𝑍𝐶. We also recall that the side
splitter theorem tells us that if a line parallel to one side of a triangle
intersects the other two sides, then the line divides those sides
proportionally. Conversely if a line splits two
sides of a triangle into equal proportions, then that line must be parallel to
the third side.
In our case, since sides 𝑋𝐷
and 𝑋𝐶 of the larger triangle 𝑋𝐷𝐶 have been divided into equal proportions,
we can apply the converse of the side splitter theorem to deduce that sides 𝐷𝐶
and 𝑌𝑍 must be parallel. Now, recall also that if a line
parallel to a side of a triangle intersects the other two sides, as does the
line 𝐼𝐽 in the diagram, then the smaller triangle created by that line, here
triangle 𝐼𝐽𝐻, is similar to the original triangle 𝐹𝐺𝐻. In our case, this means the
triangle 𝑋𝑌𝑍 is similar to triangle 𝑋𝐷𝐶.
Now, since the sides 𝐷𝐶 and
𝐴𝐵 are opposite sides of the parallelogram 𝐴𝐵𝐶𝐷, they must have equal
length. So now, we know that side 𝐷𝐶
has length 134.9 centimeters. Now, making some space, if we
extract our triangle and call the length 𝑋𝑌 lowercase 𝑥, our side length 𝑋𝐷
is equal to two 𝑥. Since our triangles 𝑋𝑌𝑍 and
𝑋𝐷𝐶 are similar, we can form an equation with the side lengths are shown such
that the proportions 𝑋𝑌 over 𝑋𝐷 and 𝑌𝑍 over 𝐷𝐶 are equal.
Now, substituting in the side
lengths we know, we have 𝑥 over two 𝑥 is equal to 𝑌𝑍 over 134.9. We can divide top and bottom on
the left by 𝑥 to give one over two is equal to 𝑌𝑍 over 134.9. Then multiplying through by
134.9, we have 𝑌𝑍 equals 134.9 over two, which evaluates to 67.45. The length 𝑌𝑍 is therefore
67.45 centimeters.
Let’s now, finish by recapping some
of the main points we’ve covered. If a line intersecting two sides of
a triangle is parallel to the remaining side, then the smaller triangle created by
this line is similar to the original triangle. The side splitter theorem tells us
that if a line parallel to one side of a triangle intersects the other two sides of
the triangle, then the line divides those sides proportionally.
The side splitter theorem extends
to include parallel lines outside a triangle. If a line outside a triangle is
parallel to one of the sides of the triangle and intersects the extensions of the
other two sides, then the line divides the extensions of those sides
proportionally. And finally, the converse of the
side splitter theorem states that if a line splits two sides of a triangle
proportionally, then that line is parallel to the remaining side.
